Problem: Solve for $x$ and $y$ using elimination. ${2x+y = 15}$ ${3x-y = 15}$
We can eliminate $y$ by adding the equations together when the $y$ coefficients have opposite signs. Add the equations together. Notice that the terms $y$ and $-y$ cancel out. $5x = 30$ $\dfrac{5x}{{5}} = \dfrac{30}{{5}}$ ${x = 6}$ Now that you know ${x = 6}$ , plug it back into $\thinspace {2x+y = 15}\thinspace$ to find $y$ ${2}{(6)}{ + y = 15}$ $12+y = 15$ $12{-12} + y = 15{-12}$ ${y = 3}$ You can also plug ${x = 6}$ into $\thinspace {3x-y = 15}\thinspace$ and get the same answer for $y$ : ${3}{(6)}{ - y = 15}$ ${y = 3}$